This problem involves using the inclusion-exclusion principle to count the number of students majoring in each subject, along with some set theory to answer the given questions. Let's solve the problem step by step.

Step 1: Define the Sets

Let:

• $C$ be the set of students majoring in Chemistry,
• $P$ be the set of students majoring in Physics,
• $M$ be the set of students majoring in Mathematics.

We are given the following information:

• $|C| = 10$ (students majoring in Chemistry),
• $|P| = 13$ (students majoring in Physics),
• $|C \cap P| = 7$ (students majoring in both Chemistry and Physics),
• $|C \cap M| = 6$ (students majoring in both Chemistry and Mathematics),
• $|P \cap M| = 8$ (students majoring in both Physics and Mathematics),
• $|C \cap P \cap M| = 5$ (students majoring in all three subjects),
• The total number of students is $20$.

Step 2: Calculate the Number of Students Majoring in Mathematics

We will use the inclusion-exclusion principle to find $|M|$, the number of students majoring in Mathematics.

$|M| = |C \cap M| + |P \cap M| - |C \cap P \cap M| + \text{(students who only major in Mathematics)}$

We can rearrange this to:

$|M| = |C \cap M| + |P \cap M| - |C \cap P \cap M| + x$

Where $x$ is the number of students who only major in Mathematics.

First, let’s figure out the total number of students in all categories using the inclusion-exclusion principle:

$|C \cup P \cup M| = |C| + |P| + |M| - |C \cap P| - |C \cap M| - |P \cap M| + |C \cap P \cap M|$

Substitute the known values:

$20 = 10 + 13 + |M| - 7 - 6 - 8 + 5$

Simplifying this:

$20 = 22 + |M| - 16$ $20 = 6 + |M|$ $|M| = 14$

Thus, the number of students majoring in Mathematics is $\boxed{14}$.

Step 3: Probability of Majoring in Exactly Two Subjects

To find the probability that a randomly selected Mathematics major is majoring in exactly two subjects, we first need to determine how many Mathematics majors are in exactly two subjects.

From the problem, the students majoring in two subjects are:

• $|C \cap M| - |C \cap P \cap M| = 6 - 5 = 1$ (students majoring in Chemistry and Mathematics but not Physics),
• $|P \cap M| - |C \cap P \cap M| = 8 - 5 = 3$ (students majoring in Physics and Mathematics but not Chemistry).

So, the total number of Mathematics majors in exactly two subjects is:

$1 + 3 = 4$

Now, the probability that a randomly selected Mathematics major is in exactly two subjects is:

$\text{Probability} = \frac{\text{Number of Mathematics majors in exactly two subjects}}{\text{Total number of Mathematics majors}} = \frac{4}{14} = \frac{2}{7}$

Thus, the probability is $\boxed{\frac{2}{7}}$.

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Do you want more details or have any questions?

Here are five related questions:

1. How many students major in only one subject?
2. What is the probability that a randomly selected Physics major also majors in Chemistry?
3. How many students major in exactly three subjects?
4. How many students major only in Mathematics?
5. What is the total number of students who major in more than one subject?

Tip: The inclusion-exclusion principle is very useful when dealing with sets and overlapping categories.